Linear System with Parameter

Here is problem 3 from the Fifth International Internet Mathematical Olympiad for Students. This is an online competition run by the Ariel University Center of Samaria, Israel.

Find all values of the parameter y for which the following system has a solution

(1)x5 + x2= yx1
(2)x1 + x3= yx2
(3)x2 + x4= yx3
(4)x3 + x5= yx4
(5)x4 + x1= yx5

For each value of y, find all the solutions.

Solution

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Copyright © 1996-2017 Alexander Bogomolny

Find all values of the parameter y for which the following system has a solution

(1)x5 + x2= yx1
(2)x1 + x3= yx2
(3)x2 + x4= yx3
(4)x3 + x5= yx4
(5)x4 + x1= yx5

For each value of y, find all the solutions.

First of all observe that, for any value of y, the system has a trivial solution: x1 = x2 = x3 = x4 = x5 = 0. The task is to find the values of y for which there is a nontrivial solution. Add up all five equations to obtain

  (y - 2)(x1 + x2 + x3 + x4 + x5) = 0.

So, either y = 2 or x1 + x2 + x3 + x4 + x5 = 0. Clearly the value y = 2 is bound to play a prominent role in the solution. Curiously, there are other y's of no lesser importance.

But, to follow up, assume first that y = 2. The system becomes

(1')x5 + x2= 2x1
(2')x1 + x3= 2x2
(3')x2 + x4= 2x3
(4')x3 + x5= 2x4
(5')x4 + x1= 2x5

Since the system is homogeneous and we are looking for a non-trivial solution, we may omit the last equation and solve the rest with the Gaussian elimination, getting subsequently,

 
- 21001
1 - 2100
01- 210
001- 21
 ⇒ 
- 21001
0- 3201
00- 431
000- 55

By the back substitution, x5 = x4 = x3 = x2 = x1, with arbitrary x1.

Now we may assume y ≠ 2 and x1 + x2 + x3 + x4 + x5 = 0. We'll replace (quite arbitrarily) the last equation with this one and employ again the Gaussian elimination.

 
11111
-y1001
1 -y100
01 -y10
001 -y1
 ⇒ 
11111
0y+1yyy+1
0 -(y+1)0 -1-1
01 -y10
001 -y1
 ⇒ 
 
11111
01 -y10
0y+1yyy+1
0 -(y+1)0 -1-1
001 -y1
 ⇒ 
11111
01 -y10
00yy-1y
00-y(y+1)y-1
001 -y1
 ⇒ 
 
11111
01 -y10
00-y(y+1)y-1
00yy-1y
00y+1 -1y+1
 ⇒ 
11111
01 -y10
00-y(y+1)y-1
000y²+y-1y²+y-1
0000y²+y-1

If y² + y - 1 ≠ 0, the last equation gives x5 = 0 and the back substitution leads to a trivial solution which was already accounted for. So let us assume that y² + y - 1 = 0. The quadratic formula gives two distinct solutions

  y1,2 = (-1 ± 5) / 2.

With the parameter y equal to one of them, the last equation implies that x5 can be chosen arbitrarily; and, from the fourth equation, the same is true of x4.

Since, for the chosen y, -y(y + 1) = -1, the third equation gives

  x3 = yx4 - x5

From the second equation,

  x2 = -y(x4 + x5)

and, from the first equation,

  x1 = -x4 + yx5.

Thus the system has three sets of solutions:

  1. A trivial solution for any y,

  2. A 1-parameter solution for y = 2,

  3. A 2-parameter solution for y = (-1 ± 5) / 2.

The latter family, perhaps surprisingly, includes the golden ratio.

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Copyright © 1996-2017 Alexander Bogomolny

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